/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 According to the Eurostat Statis... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to the Eurostat Statistics Database, Spain accounts for \(19 \%\) of the total fishes caught in the European Union (EU). Assume that we randomly sample two fishes caught in the \(\mathrm{EU}\). a. If a fish is caught in Spain, record \(Y ;\) if not, record \(N\). List all possible sequences of \(Y\) and \(N\). b. For each sequence, find by hand the probability that it will occur, assuming each outcome is independent. c. What is the probability that neither of the two randomly selected fishes have been caught in Spain? d. What is the probability that exactly one out of the two fishes has been caught in Spain? e. What is the probability that both have been caught in Spain?

Short Answer

Expert verified
a) The possible sequences are YY, YN, NY, NN. \nb) \(P(YY)=0.0361\), \(P(YN)=0.1531\), \(P(NY)=0.1531\), \(P(NN)=0.6561\). \nc) The probability that none of the fish comes from Spain is \(0.6561\). \nd) The probability that exactly one fish comes from Spain is \(0.3062\). \ne) The probability that both fish come from Spain is \(0.0361\).

Step by step solution

01

List all possible sequences

The possible sequences are: YY, YN, NY, NN. The sequences YY and NN represent both fish caught in Spain and both fish not caught in Spain respectively, while YN and NY represent one fish caught in Spain and the other not caught in Spain.
02

Calculate probability for each sequence

Since these are independent events, the probability of each sequence occurring is the product of the probabilities of each individual event. The probability a fish is caught in Spain is \(0.19\) or \(19\%\), hence: \n- \(P(YY) = P(Y) \cdot P(Y) = 0.19 \cdot 0.19 = 0.0361\) \n- \(P(YN) = P(Y) \cdot P(N) = 0.19 \cdot (1-0.19) = 0.1531\), \n- \(P(NY) = P(N) \cdot P(Y) = (1-0.19) \cdot 0.19 = 0.1531\) \n- \(P(NN) = P(N) \cdot P(N) = (1-0.19) \cdot (1-0.19) = 0.6561\)
03

Calculate probability for stated scenarios

c) The probability that neither of the two randomly selected fishes have been caught in Spain is the \(P(NN)\), which is \(0.6561\). \nd) The probability that exactly one out of the two fishes has been caught in Spain is the sum of \(P(YN)\) and \(P(NY)\), which is \(0.1531 + 0.1531 = 0.3062\). \ne) The probability that both have been caught in Spain is \(P(YY)\), which is \(0.0361\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
Understanding independent events is crucial when learning about probability. In statistics, two events are considered independent if the occurrence of one event does not influence the occurrence of the other. For instance, flipping a coin and rolling a die are independent events because the result of the coin flip doesn't affect the outcome of the die roll.

In the case of the fish caught in Spain, the selection of each fish can be seen as an independent event. That is, catching one fish from Spain does not affect the chances of catching another fish from Spain. Consequently, when we calculate the probabilities of different sequences such as 'YY', 'YN', 'NY', and 'NN', we can use the rule of product for independent events. The rule states that the probability of both events occurring is the product of their individual probabilities, symbolized mathematically as:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
Calculating Probabilities
Calculating probabilities is a fundamental aspect of statistics that entails determining the likelihood of various outcomes. To calculate the probability of an event, one generally divides the number of favorable outcomes by the total number of possible outcomes. When events are independent, as discussed earlier, the calculation entails the multiplication of their individual probabilities.

For a visual breakdown, if we imagine a bag with 100 marbles, of which 19 are blue (representing fishes caught in Spain), and 81 are red (representing fishes caught elsewhere in the EU), the probability of drawing a blue marble is \( \frac{19}{100} = 0.19 \). If we are to draw two marbles one after the other, the probabilities of all possible outcomes ('YY', 'YN', 'NY', 'NN') can be calculated using the individual probabilities and the rule of product for independent events. This is especially useful in scenarios that involve repeated experiments or samplings, such as drawing multiple marbles from a bag or catching multiple fishes.
Binomial Probability
The binomial probability is a specific type of probability that applies in situations where there are two possible outcomes for a series of independent events or trials, often labeled as 'success' and 'failure'. The characteristics of a binomial experiment include a fixed number of trials, only two possible outcomes per trial, each trial is independent, and the probability of success remains constant throughout the trials.

In our fish example, catching a fish from Spain can be viewed as a 'success' (Y) and not catching one as a 'failure' (N). Finding the probability that both in two attempts have been caught from Spain, or that exactly one or none have been caught from Spain, directly relates to the binomial probability distribution. The probability of exactly k successes in n trials in a binomial distribution is given by the formula:
\[ P(k; n, p) = \binom{n}{k} p^k (1 - p)^{(n - k)} \]
where \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose k successes from n trials, p is the probability of success on any given trial, and 1-p is the probability of failure. By applying this formula, you can easily find the probabilities for each of our scenarios (c, d, and e) in the original exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Extreme Negative z-Scores For each question, find the area to the right of the given z-score in a standard Normal distribution. In this question, round your answers to the nearest \(0.000 .\) Include an appropriately labeled sketch of the \(N(0,1)\) curve. a. \(z=-4.00\) b. \(z=-8.00\) c. \(z=-30.00\) d. If you had the exact probability for these right proportions, which would be the largest and which would be the smallest? e. Which is equal to the area in part b: the area below (to the left of) \(z=8.00\) or the area above (to the right of) \(z=8.00 ?\)

Use a table or technology to answer each question. Include an appropriately labeled sketch of the Normal curve for each part. Shade the appropriate region. a. Find the area to the left of a \(z\) -score of \(-0.50\). h. Find the area to the right of a \(z\) -score of \(-0.50\).

The homicide clearance rate in Florida is \(60 \%\). A crime is cleared when an arrest is made, a crime is charged, and the case is referred to a court. a. What is the probability that exactly 6 out of 10 independent homicides are cleared? b. Without doing a calculation, state whether the probability that 6 or more out of 10 homicides are cleared will be larger or smaller than the answer to part a? Why? c. What is the probability that 6 or more out of 10 independent homicides are cleared?

a. Number of months in a calendar year b. The time taken by carth to complete one revolution around the sun

A survey shows that in one year, the average number of bilingual doctors in a hospital was \(42 .\) Assume that the standard deviation is 12 and the number of bilingual doctors in a hospital is normally distributed. Include an appropriately labeled and shaded Normal curve for each part. a. What percentage of hospitals have between 40 and 45 bilingual doctors? b. What percentage of hospitals have between 45 and 50 bilingual doctors?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.